Question

Can someone help me find the range of this function?

h(x) = -2x^2 + 12x + 5

I know the domain is all real numbers, but I'm confused on the range. Thanks.

Answer 1

h(x) = -2x^2 + 12x + 5
h(x) = -2(x^2-6x-2.5)
h(x) = -2(x-3)^2 -11.5
so h is 3, k is -11.5 and a is -2
so range goes from -infinite to -11.5

Answer 2

Hmm, the answer is saying the range is from -infinity to 23

Answer 3

h(x) = -2x^2 + 12x + 5


h(x) = -2(x^2 - 6x - 5/2)


h(x) = -2(x^2 - 6x + 9 - 9 - 5/2)


h(x) = -2((x^2 - 6x + 9) - 9 - 5/2)


h(x) = -2((x - 3)^2 - 9 - 5/2)


h(x) = -2((x - 3)^2 - 23/2)


h(x) = -2(x - 3)^2 - 2(-23/2)


h(x) = -2(x - 3)^2 + 23


So the range is (-inf, 23]

Answer 4

oups, sry

Answer 5

-infinity to 23

How so i think it is all real numbers.

Answer 6

If you graph the function, you'll see that it peaks at the point (3,23). So only y values that are equal to 23 or less than it will be part of the range.